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Let $I$ be the open interval $(0, 1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Exercise 8.16 Let $E=L^p(I)$ with $1 \leq p<\infty$. Consider the unbounded operator $A: D(A) \subset E \rightarrow E$ defined by $$ D(A)=\left\{u \in W^{1, p}(I) : u(0)=0\right\} \quad \text { and } \quad A u=u^{\prime} . $$

  1. Check that $D(A)$ is dense in $E$ and that $A$ is closed (i.e., its graph $G(A)$ is closed in $E \times E$).
  2. Determine $R(A)$ and $N(A)$.
  3. Compute $A^{*}$. Check that $D\left(A^{*}\right)$ is dense in $E^{*}=L^{p'}(I)$ when $1<p<\infty$, but $D\left(A^{*}\right)$ is not dense in $E^{*}=L^{\infty}(I)$ when $p=1$. Here $p'$ is the Hölder conjugate of $p$.

There are possibly subtle mistakes that I could not recognize in my below attempt. Could you please have a check on it?

  1. We have that $C_c^1 (I) \subset W_0^{1, p} (I) \subset D(A)$ and that $C^\infty_c (I)$ is dense in $E$. Then $D(A)$ is dense in $E$. Let's prove that $G(A)$ is closed in $E \times E$. Let $(u_n, u'_n)$ be a sequence in $G(A)$ such that $(u_n, u'_n) \to (u, v)$ in $E \times E$. First, we prove that $u \in W^{1, p}(I)$ with $u' =v$. Indeed, $$ \int_I u_n \varphi' = - \int_I u_n' \varphi, \quad \varphi \in C^1_c (I), $$ and at the limit $$ \int_I u \varphi' = - \int_I v \varphi, \quad \varphi \in C^1_c (I). $$

It is clear that $u_n \to u$ in $W^{1, p}(I)$. By Theorem 8.8 (in the same book), $W^{1, p}(I) \subset L^{\infty}(I)$ with continuous injection. Then $u(0)=0$ and thus $u \in G(A)$. This completes the proof.

Akira
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1 Answers1

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Note that showing that $A$ is closed, it is equivalent to showing that $$ G(A)=\{(u,Au): u\in D(A)\} $$ is closed. In fact, let $u_n \in D(A)$ and $u_n \to u$, $Au_n = u_n' \to g$ in $L^p(I)$. Then we show that $u \in L^p(I)$ and $u' = g$. In fact, $$ \bigg|u_n(t)-\int_0^tg(s)ds\bigg| = \bigg|\int_0^t (u_n'(s)-g(s)) ds\bigg|\le t^{\frac1{p'}}\bigg(\int_0^t |u_n'(s)-g(s)|^p ds\bigg)^{\frac1p}\le\|u_n'-g\|_{L^p} $$ and hence $$ \bigg(\int_I\bigg|u_n(t)-\int_0^tg(s)ds\bigg|^pdt\bigg)^{\frac1p}\le\|u_n'-g\|_{L^p}\to 0\text{ as }n\to\infty. $$ This implies $u\in L^p(I)$ and $$ u(t)=\int_0^tg(s)ds \text{ or }u'=g, u(0)=0.$$ This proves that $A$ is closed.

xpaul
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